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Proof that f(z) is in no point analytic

  1. Sep 4, 2014 #1
    1. The problem statement, all variables and given/known data

    I have to proof for a homework assignment that f(z) = z^(n)(z*)^(-m) where n, m ≥ 1 (are natural numbers) is in no point analytic.

    2. Relevant equations

    binomial function

    3. The attempt at a solution

    I found this wikipedia page, were they state that:
    a complex-valued function ƒ of a complex variable z is said to be analytic at a if in some open disk centered at a it can be expanded as a convergent power series.

    for wich i think i need to use the binomial function.

    But i'm not sure how to do this for the given function f(z)
     
    Last edited: Sep 4, 2014
  2. jcsd
  3. Sep 4, 2014 #2
    (z*) is the complex conjugate


    If i look at the Lim (z->0) z/z* i would say that it does not exist because if i look let z approach 0 on the real axis i get 1, and if i let z approach 0 on the imaginary axis i get -1.

    but now i think i have to look at the lim(z->0) z^n/(z*)^m and am not sure how to approach the powers.

    thanx
     
    Last edited: Sep 4, 2014
  4. Sep 4, 2014 #3

    vela

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  5. Sep 4, 2014 #4
    So the function is analytic in a domain if it is differentiable at all points of the domain.
     
  6. Sep 4, 2014 #5
    There is yet another criterion based on the so called Wirtinger derivatives

    see: http://wcherry.math.unt.edu/math5410/wirtinger.pdf

    Roughly, if [itex] \frac{\partial f(z,\overline{z})}{\partial \overline{z}} \neq 0[/itex], then [itex]f[/itex] does not satisfy the Cauchy-Riemann equations, and thus cannot be analytic.

    Edit: Here [itex]\overline{z}[/itex] is the conjugate of [itex]z[/itex]
     
    Last edited: Sep 4, 2014
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